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# I can't solve it, simple maths

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Question:

How mnay 3-digit numbers can be constructed from the digiits 1, 2, 3, 4, 5, 6, and 7 if each digit may be used:

a. As often as desired: 343

b. Only once: 210

c. Once only and the number must be odd?

I couldn't do C, I'd assume that you divide 210 by 2 to get 105. But the answer is 120, what am I missing out here?

Thanks.

Guest Sep 10, 2017
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#1
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Hi Guest,

I really like that you have explained what you did :)

Lets take a look.

Question:

How mnay 3-digit numbers can be constructed from the digiits 1, 2, 3, 4, 5, 6, and 7 if each digit may be used:

a. As often as desired: 343

Lets do units*tens* hundreds

7*7*7

b. Only once: 210

There is no zero so any digit can be anywhere.

7*6*5

I couldn't do C, I'd assume that you divide 210 by 2 to get 105. But the answer is 120, what am I missing out here?

c. Once only and the number must be odd?

The units this time can be 1 or 3 or 5 or 7  That is 4 possibilities

So that is  4*6*5 = 120

does that help?

Melody  Sep 10, 2017
edited by Melody  Sep 10, 2017
#2
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Thanks.

Guest Sep 10, 2017

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